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00 mL of 0.0120 M Pb(NO2)2 are mixed with 3.00 mL 0.0300 M KI. And diluted to a final volume of 10.00 mL with KNO3 solvent. A yellow precipitate of...

5.00 mL of 0.0120 M Pb(NO2)2 are mixed with 3.00 mL 0.0300 M KI. And diluted to a final volume of 10.00 mL with KNO3 solvent. A yellow precipitate of PbI2 forms. The precipitate is separated from the supernatant solution which is analyzed for [I–]. The concentration of I– in the test solution is calculated from the graph to be 1.82 x 10–3 M.

Calculate the following:

a. Initial [Pb+2] (after mixing but before precipitating)
b. Initial [I–] (after mixing but before precipitating)
c. Equilibrium [I–] = (graph [I–] x 10.0 mL/3.00 mL)
d. [I–] reacted to form precipitate
e. [Pb+2] reacted to form precipitate
f. Equilibrium [Pb+2]
g. Ksp of PbI2
h. Tube #1 is analyzed by taking 3.00 mL of the saturated solution of PbI2 and treating it with acid and KNO2 and diluted to 10.00 mL. This diluted solution is then analyzed in a spectrophotometer and found to have [I–] of 1.59 x 10–3 M using the regression equation from the graph. Calculate the Ksp of PbI2 using this data.

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Chemistry-8201341.doc

5.00 mL of 0.0120 M Pb(NO2)2 are mixed with 3.00 mL 0.0300 M KI. And diluted to a final volume of
10.00 mL with KNO3 solvent. A yellow precipitate of PbI2 forms. The precipitate is separated from...

Sign up to view the full answer

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